We are given two arrays
A
andB
of words. Each word is a string of lowercase letters.Now, say that word
b
is a subset of worda
if every letter inb
occurs ina
, including multiplicity. For example,"wrr"
is a subset of"warrior"
, but is not a subset of"world"
.Now say a word
a
fromA
is universal if for everyb
inB
,b
is a subset ofa
.Return a list of all universal words in
A
. You can return the words in any order.
Example 1:
Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]Example 2:
Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]Example 3:
Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]Example 4:
Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]Example 5:
Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]
Note:
1 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i]
andB[i]
consist only of lowercase letters.- All words in
A[i]
are unique: there isn'ti != j
withA[i] == A[j]
.
Approach #1: String. [Java]
class Solution {
public List<String> wordSubsets(String[] A, String[] B) {
int[] init = new int[26], temp;
for (String b : B) {
temp = counter(b);
for (int i = 0; i < 26; ++i) {
init[i] = Math.max(init[i], temp[i]);
}
}
List<String> ret = new ArrayList<>();
for (String a : A) {
temp = counter(a);
int i = 0;
for (i = 0; i < 26; ++i) {
if (temp[i] < init[i]) break;
}
if (i == 26) ret.add(a);
}
return ret;
} public int[] counter(String b) {
int[] temp = new int[26];
for (int i = 0; i < b.length(); ++i) {
temp[b.charAt(i)-'a']++;
}
return temp;
}
}
Reference:
https://leetcode.com/problems/word-subsets/discuss/175854/C%2B%2BJavaPython-Straight-Forward